Prove $Bbb Z/4Bbb Z times Bbb Z/6Bbb Z to Bbb Z/4Bbb Z times Bbb Z/3Bbb Z$ is a homomorphism given by $([x],...
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Prove $Bbb Z/4Bbb Z times Bbb Z/6Bbb Z to Bbb Z/4Bbb Z times Bbb Z/3Bbb Z$ is a homomorphism given by $([x], [y]) to([x+2y], [y])$.
I don't know how to prove this because $Bbb Z/6Bbb Z$ is NOT homomorphic to $Bbb Z/3Bbb Z$ right? Then, how can $Bbb Z/4Bbb Z times Bbb Z/6Bbb Z to Bbb Z/4Bbb Z times Bbb Z/3Bbb Z$ be homomorphic?
abstract-algebra
edited Nov 19 at 5:52
Tianlalu
2,9901936
asked Nov 19 at 4:56
david D
875
add a comment |
up vote
0
down vote
favorite
Prove $Bbb Z/4Bbb Z times Bbb Z/6Bbb Z to Bbb Z/4Bbb Z times Bbb Z/3Bbb Z$ is a homomorphism given by $([x], [y]) to([x+2y], [y])$.
I don't know how to prove this because $Bbb Z/6Bbb Z$ is NOT homomorphic to $Bbb Z/3Bbb Z$ right? Then, how can $Bbb Z/4Bbb Z times Bbb Z/6Bbb Z to Bbb Z/4Bbb Z times Bbb Z/3Bbb Z$ be homomorphic?
abstract-algebra
edited Nov 19 at 5:52
Tianlalu
2,9901936
asked Nov 19 at 4:56
david D
875
There is always a trivial homomorphism between groups.
– CyclotomicField
Nov 19 at 5:22
That isn't well-defined.
– Lord Shark the Unknown
Nov 19 at 5:35
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Prove $Bbb Z/4Bbb Z times Bbb Z/6Bbb Z to Bbb Z/4Bbb Z times Bbb Z/3Bbb Z$ is a homomorphism given by $([x], [y]) to([x+2y], [y])$.
I don't know how to prove this because $Bbb Z/6Bbb Z$ is NOT homomorphic to $Bbb Z/3Bbb Z$ right? Then, how can $Bbb Z/4Bbb Z times Bbb Z/6Bbb Z to Bbb Z/4Bbb Z times Bbb Z/3Bbb Z$ be homomorphic?
abstract-algebra
edited Nov 19 at 5:52
Tianlalu
2,9901936
asked Nov 19 at 4:56
david D
875
Prove $Bbb Z/4Bbb Z times Bbb Z/6Bbb Z to Bbb Z/4Bbb Z times Bbb Z/3Bbb Z$ is a homomorphism given by $([x], [y]) to([x+2y], [y])$.
I don't know how to prove this because $Bbb Z/6Bbb Z$ is NOT homomorphic to $Bbb Z/3Bbb Z$ right? Then, how can $Bbb Z/4Bbb Z times Bbb Z/6Bbb Z to Bbb Z/4Bbb Z times Bbb Z/3Bbb Z$ be homomorphic?
abstract-algebra
abstract-algebra
edited Nov 19 at 5:52
Tianlalu
2,9901936
asked Nov 19 at 4:56
david D
875
edited Nov 19 at 5:52
Tianlalu
2,9901936
asked Nov 19 at 4:56
david D
875
edited Nov 19 at 5:52
Tianlalu
2,9901936
edited Nov 19 at 5:52
Tianlalu
2,9901936
edited Nov 19 at 5:52
Tianlalu
2,9901936
2,9901936
asked Nov 19 at 4:56
david D
875
asked Nov 19 at 4:56
david D
875
asked Nov 19 at 4:56
david D
875
875
There is always a trivial homomorphism between groups.
– CyclotomicField
Nov 19 at 5:22
That isn't well-defined.
– Lord Shark the Unknown
Nov 19 at 5:35
add a comment |
There is always a trivial homomorphism between groups.
– CyclotomicField
Nov 19 at 5:22
That isn't well-defined.
– Lord Shark the Unknown
Nov 19 at 5:35
There is always a trivial homomorphism between groups.
– CyclotomicField
Nov 19 at 5:22
There is always a trivial homomorphism between groups.
– CyclotomicField
Nov 19 at 5:22
That isn't well-defined.
– Lord Shark the Unknown
Nov 19 at 5:35
That isn't well-defined.
– Lord Shark the Unknown
Nov 19 at 5:35
add a comment |
1 Answer
1
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oldest
votes
up vote
1
down vote
accepted
It's enough to prove it for each component.
begin{align}
&varphi_{11}:Bbb Z/4Bbb Z to Bbb Z/4Bbb Z&
&x+4Bbb Zmapsto x+4Bbb Z\
&varphi_{12}:Bbb Z/6Bbb Z to Bbb Z/4Bbb Z&
&y+6Bbb Zmapsto 2y+4Bbb Z\
&varphi_{21}:Bbb Z/4Bbb Z to Bbb Z/3Bbb Z&
&x+4Bbb Zmapsto 0+3Bbb Z\
&varphi_{22}:Bbb Z/6Bbb Z to Bbb Z/3Bbb Z&
&y+6Bbb Zmapsto y+3Bbb Z\
end{align}
In particular, $varphi_{22}$ is the only group homomorphism making the following diagram commutative$require{AMScd}$:
begin{CD}
Bbb Z@>ymapsto y+3Bbb Z>>Bbb Z/3Bbb Z\
@VVV @|\
Bbb Z/6Bbb Z@>>varphi_{22}>Bbb Z/3Bbb Z
end{CD}
answered Nov 19 at 8:18
Fabio Lucchini
7,81311326
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
It's enough to prove it for each component.
begin{align}
&varphi_{11}:Bbb Z/4Bbb Z to Bbb Z/4Bbb Z&
&x+4Bbb Zmapsto x+4Bbb Z\
&varphi_{12}:Bbb Z/6Bbb Z to Bbb Z/4Bbb Z&
&y+6Bbb Zmapsto 2y+4Bbb Z\
&varphi_{21}:Bbb Z/4Bbb Z to Bbb Z/3Bbb Z&
&x+4Bbb Zmapsto 0+3Bbb Z\
&varphi_{22}:Bbb Z/6Bbb Z to Bbb Z/3Bbb Z&
&y+6Bbb Zmapsto y+3Bbb Z\
end{align}
In particular, $varphi_{22}$ is the only group homomorphism making the following diagram commutative$require{AMScd}$:
begin{CD}
Bbb Z@>ymapsto y+3Bbb Z>>Bbb Z/3Bbb Z\
@VVV @|\
Bbb Z/6Bbb Z@>>varphi_{22}>Bbb Z/3Bbb Z
end{CD}
answered Nov 19 at 8:18
Fabio Lucchini
7,81311326
add a comment |
up vote
1
down vote
accepted
It's enough to prove it for each component.
begin{align}
&varphi_{11}:Bbb Z/4Bbb Z to Bbb Z/4Bbb Z&
&x+4Bbb Zmapsto x+4Bbb Z\
&varphi_{12}:Bbb Z/6Bbb Z to Bbb Z/4Bbb Z&
&y+6Bbb Zmapsto 2y+4Bbb Z\
&varphi_{21}:Bbb Z/4Bbb Z to Bbb Z/3Bbb Z&
&x+4Bbb Zmapsto 0+3Bbb Z\
&varphi_{22}:Bbb Z/6Bbb Z to Bbb Z/3Bbb Z&
&y+6Bbb Zmapsto y+3Bbb Z\
end{align}
In particular, $varphi_{22}$ is the only group homomorphism making the following diagram commutative$require{AMScd}$:
begin{CD}
Bbb Z@>ymapsto y+3Bbb Z>>Bbb Z/3Bbb Z\
@VVV @|\
Bbb Z/6Bbb Z@>>varphi_{22}>Bbb Z/3Bbb Z
end{CD}
answered Nov 19 at 8:18
Fabio Lucchini
7,81311326
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
It's enough to prove it for each component.
begin{align}
&varphi_{11}:Bbb Z/4Bbb Z to Bbb Z/4Bbb Z&
&x+4Bbb Zmapsto x+4Bbb Z\
&varphi_{12}:Bbb Z/6Bbb Z to Bbb Z/4Bbb Z&
&y+6Bbb Zmapsto 2y+4Bbb Z\
&varphi_{21}:Bbb Z/4Bbb Z to Bbb Z/3Bbb Z&
&x+4Bbb Zmapsto 0+3Bbb Z\
&varphi_{22}:Bbb Z/6Bbb Z to Bbb Z/3Bbb Z&
&y+6Bbb Zmapsto y+3Bbb Z\
end{align}
In particular, $varphi_{22}$ is the only group homomorphism making the following diagram commutative$require{AMScd}$:
begin{CD}
Bbb Z@>ymapsto y+3Bbb Z>>Bbb Z/3Bbb Z\
@VVV @|\
Bbb Z/6Bbb Z@>>varphi_{22}>Bbb Z/3Bbb Z
end{CD}
answered Nov 19 at 8:18
Fabio Lucchini
7,81311326
It's enough to prove it for each component.
begin{align}
&varphi_{11}:Bbb Z/4Bbb Z to Bbb Z/4Bbb Z&
&x+4Bbb Zmapsto x+4Bbb Z\
&varphi_{12}:Bbb Z/6Bbb Z to Bbb Z/4Bbb Z&
&y+6Bbb Zmapsto 2y+4Bbb Z\
&varphi_{21}:Bbb Z/4Bbb Z to Bbb Z/3Bbb Z&
&x+4Bbb Zmapsto 0+3Bbb Z\
&varphi_{22}:Bbb Z/6Bbb Z to Bbb Z/3Bbb Z&
&y+6Bbb Zmapsto y+3Bbb Z\
end{align}
In particular, $varphi_{22}$ is the only group homomorphism making the following diagram commutative$require{AMScd}$:
begin{CD}
Bbb Z@>ymapsto y+3Bbb Z>>Bbb Z/3Bbb Z\
@VVV @|\
Bbb Z/6Bbb Z@>>varphi_{22}>Bbb Z/3Bbb Z
end{CD}
answered Nov 19 at 8:18
Fabio Lucchini
7,81311326
answered Nov 19 at 8:18
Fabio Lucchini
7,81311326
answered Nov 19 at 8:18
Fabio Lucchini
7,81311326
answered Nov 19 at 8:18
Fabio Lucchini
7,81311326
7,81311326
add a comment |
add a comment |
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There is always a trivial homomorphism between groups.
– CyclotomicField
Nov 19 at 5:22
That isn't well-defined.
– Lord Shark the Unknown
Nov 19 at 5:35